我不希望我的用户尝试下载任何东西,除非他们连接了Wi-Fi。然而,我似乎只能判断是否启用了Wi-Fi,但他们仍然可能有3G连接。

android.net.wifi.WifiManager m = (WifiManager) getSystemService(WIFI_SERVICE);
android.net.wifi.SupplicantState s = m.getConnectionInfo().getSupplicantState();
NetworkInfo.DetailedState state = WifiInfo.getDetailedStateOf(s);
if (state != NetworkInfo.DetailedState.CONNECTED) {
    return false;
}

然而,这种状态并不是我所期望的。即使Wi-Fi是连接的,我得到OBTAINING_IPADDR作为状态。


当前回答

这个问题有点老了,但我用的就是这个。要求最低api级别21也考虑到废弃的Networkinfo api。

boolean isWifiConn = false;
    ConnectivityManager connMgr = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
        Network network = connMgr.getActiveNetwork();
        if (network == null) return false;
        NetworkCapabilities capabilities = connMgr.getNetworkCapabilities(network);
        if(capabilities != null && capabilities.hasTransport(NetworkCapabilities.TRANSPORT_WIFI)){
            isWifiConn = true;
            Toast.makeText(context,"Wifi connected Api >= "+Build.VERSION_CODES.M,Toast.LENGTH_LONG).show();
        }else{
            Toast.makeText(context,"Wifi not connected Api >= "+Build.VERSION_CODES.M,Toast.LENGTH_LONG).show();
        }
    } else {
        for (Network network : connMgr.getAllNetworks()) {
            NetworkInfo networkInfo = connMgr.getNetworkInfo(network);
            if (networkInfo.getType() == ConnectivityManager.TYPE_WIFI && networkInfo.isConnected()) {
                isWifiConn = true;
                Toast.makeText(context,"Wifi connected ",Toast.LENGTH_LONG).show();
                break;
            }else{
                Toast.makeText(context,"Wifi not connected ",Toast.LENGTH_LONG).show();
            }
        }
    }
    return isWifiConn;

其他回答

下面是我在我的应用程序中使用的实用方法:

public static boolean isDeviceOnWifi(final Context context) {
        ConnectivityManager connManager = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
        NetworkInfo mWifi = connManager.getNetworkInfo(ConnectivityManager.TYPE_WIFI);
        return mWifi != null && mWifi.isConnectedOrConnecting();
}

使用WifiManager你可以做到:

WifiManager wifi = (WifiManager) getSystemService (Context.WIFI_SERVICE);
if (wifi.getConnectionInfo().getNetworkId() != -1) {/* connected */}

方法getNeworkId只有在没有连接到网络时才返回-1;

试试这个方法。

public boolean isInternetConnected() {
    ConnectivityManager conMgr = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);
    boolean ret = true;
    if (conMgr != null) {
        NetworkInfo i = conMgr.getActiveNetworkInfo();

        if (i != null) {
            if (!i.isConnected()) {
                ret = false;
            }

            if (!i.isAvailable()) {
                ret = false;
            }
        }

        if (i == null)
            ret = false;
    } else
        ret = false;
    return ret;
}

这种方法将有助于找到互联网连接是否可用。

许多答案使用了废弃的代码,或者在更高API版本上可用的代码。现在我用这样的东西

ConnectivityManager connectivityManager = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
        if(connectivityManager != null) {
            for (Network net : connectivityManager.getAllNetworks()) {
                NetworkCapabilities nc = connectivityManager.getNetworkCapabilities(net);
                if (nc != null && nc.hasTransport(NetworkCapabilities.TRANSPORT_WIFI)
                        && nc.hasCapability(NetworkCapabilities.NET_CAPABILITY_INTERNET))
                    return true;
            }
        }
        return false;
ConnectivityManager manager = (ConnectivityManager) getSystemService(CONNECTIVITY_SERVICE);
boolean is3g = manager.getNetworkInfo(
                  ConnectivityManager.TYPE_MOBILE).isConnectedOrConnecting();
boolean isWifi = manager.getNetworkInfo(
                    ConnectivityManager.TYPE_WIFI).isConnectedOrConnecting();

Log.v("", is3g + " ConnectivityManager Test " + isWifi);
if (!is3g && !isWifi) {
    Toast.makeText(
        getApplicationContext(),
        "Please make sure, your network connection is ON ",
        Toast.LENGTH_LONG).show();
}
else {
    // Put your function() to go further;
}